Question

A certain quantity of real gas occupies a volume of 30.15 dm3 at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300K (When its compressibility factor is 1.4) is _____ x 10-4 dm3 (Nearest integer)

Answer 1

Correct Answer: 392

We can use the relation for the compressibility factor $Z$ of a real gas: $$Z = \frac{PV}{nRT}.$$ Given the compressibility factor, we can modify the ideal gas equation: $$V = \frac{ZnRT}{P}.$$ For two different states of the gas, we can set up the following equation: $$\frac{V_2}{V_1} = \frac{Z_2 P_1 T_1}{Z_1 P_2 T_2}.$$ Given values:

• $V_1 = 0.15 \, \text{dm}^3$
• $P_1 = 100 \, \text{atm}$
• $T_1 = 500 \, \text{K}$
• $Z_1 = 1.07$
• $P_2 = 300 \, \text{atm}$
• $T_2 = 300 \, \text{K}$
• $Z_2 = 1.4$

Substitute these values into the equation: $$V_2 = V_1 \times \frac{Z_2 P_1 T_1}{Z_1 P_2 T_2}.$$ $$V_2 = 0.15 \times \frac{1.4 \times 100 \times 500}{1.07 \times 300 \times 300}.$$ $$V_2 \approx 0.15 \times \frac{70000}{96300} \approx 0.1089 \, \text{dm}^3.$$ $$V_2 \approx 108.9 \times 10^{-3} \, \text{dm}^3.$$ Thus, the volume of the gas at 300 atm and 300 K is approximately $108.9 \times 10^{-3} \, \text{dm}^3$